SOLUTION: Suppose at^2+ 5t + 4 > 0 for every real number t. Show that a >25/16.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Suppose at^2+ 5t + 4 > 0 for every real number t. Show that a >25/16.      Log On


   

Question 942945: Suppose
at^2+ 5t + 4 > 0
for every real number t. Show that a >25/16.
Found 2 solutions by richard1234, josgarithmetic:
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
The given quadratic never intersects the x-axis so it has two non-real roots. Also, it is above the x-axis so a > 0.

Therefore the discriminant must be < 0.




Rearranging, this is equivalent to .

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Critical values are the roots.

roots:
t=%28-5%2B-+sqrt%2825-4a%2A4%29%29%2F%282a%29 and you want the discriminant to be real, not complex with any imaginaries.

25-16a%3E=0, the requirement for the discriminant.
25%3E=16a;
Your really want 25%3E16a
25%2F16%3Ea
different from what you expected.